How do you solve by substitution #x+14y=84# and #2x-7= -7#?

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Answer 1 · 2015-07-11T23:29:41

#x=0#
#y=6#

There are a few ways to find #x# and #y#.

Well, we could multiply by #-2# to both sides of the equal sign of

#x+14y=84#

to get: #-2x-28y=-168#

Then, add what we just obtained to #2x-7=-7#

#2x-7=-7#
#-2x-28y=-168#

which gives us: #-7-28y=-175#

Add #7# to both sides: #-28y=-168#

Divide both sides by #-28# to get:

#y=6#

Now that we have #y#, plug it into #x+14y=84#

#x+14y=84#
#x+14(6)=84#
#x+84=84#
#x=0#

So, #x=0# and #y=6#